The distance between a pair of skew diagonals on two adjacent faces of a cube.

Question

Say we're interested in the distance between the diagonals $u=(0,0,0)+(1,0,1)t$ and $v=(0,0,1)+(1,1,0)s$ of a unit cube.

The standard formula for the distance between two skew lines $$d=|\mathbf n\cdot (\mathbf a - \mathbf b) |$$ gives $\frac{1}{\sqrt{3}}$. That's likely the answer I am seeking. But, since my original, intuitve approach conflicts with this, I don't want to give up on it.

I considered the distance from the center of a face to the center of an adjacent face. That distance is $\frac{\sqrt{2}}{2}$. The distance from a cube vertex to the center of an "adjacent" face is also $\frac{\sqrt{2}}{2}$ (for example, from $(1,0,1)$ to $(\frac{1}{2},0,1)$ or from $(0,0,1)$ to $(\frac{1}{2},0,\frac{1}{2})$). Since the distance between the two lines is unique, I know then that the distance I want is less than that.

Where is that unique segment, perpedicular to both diagonals?

Answer 1

Edited to get the picture right.

I don't have much to add to André Nicolas' answer, except a picture I quickly drew with Mathematica.

With a little imagination one can see that the minimal line indeed makes right angles with either diagonal.

The Mathematica code used to generate the picture is

s = 2/3; t = 1/3;

Graphics3D[{Line[{{0, 0, 0}, {1, 0, 0}, {1, 1, 0}, {0, 1, 0}, {0, 0, 0}}],
  Line[{{0, 0, 1}, {1, 0, 1}, {1, 1, 1}, {0, 1, 1}, {0, 0, 1}}],
  Line[{{0, 0, 0}, {0, 0, 1}}],
  Line[{{0, 1, 0}, {0, 1, 1}}],
  Line[{{1, 1, 0}, {1, 1, 1}}],
  Line[{{1, 0, 0}, {1, 0, 1}}],
  {Red, Line[{{0, 0, 0}, {1, 0, 1}}],
  Line[{{0, 0, 1}, {1, 1, 1}}]},
  {Blue, PointSize[Large], Point[{s, 0, s}], Point[{t, t, 1}], 
   Line[{{s, 0, s}, {t, t, 1}}]}
  }, Boxed -> False]

Answer 2

We want to choose $s$ and $t$ so that the distance from $(s,s,1)$ to $(t,0,t)$ is minimized. The square of the distance is $$(s-t)^2+s^2+(1-t)^2,$$ which simplifies to $2s^2+2t^2 -2st -2t+1$. To minimize, set the partial derivatives equal to $0$. We get $4s-2t=0$, $4t-2s-2=0$.

When we solve the system of two equations, we get $s=1/3$, $t=2/3$. This identifies the points of nearest approach on each line.

If we substitute in the expression for square of distance, we find that the minimum distance is indeed $\dfrac{1}{\sqrt{3}}$.

There is undoubtedly a much nicer geometric approach.

Answer 3

A geometric approach: we only use the concept of orthogonality.

The two diagonals lie respectively on the lines $\ell_1,\ell_2$ which we write as affine spaces as $$\ell_1=\langle u_1\rangle, \quad \ell_2= \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}+\langle u_2\rangle $$ where $u_1=(1, 0, 1)$ and $u_2=(1,1,0)$. Now let $v$ be the generic vector connecting the two lines, $$v=P_s-P_t=\begin{pmatrix} s-t\\ s\\ 1-t \end{pmatrix}$$ Imposing perpendicularity $v\perp u_1$ and $v\perp u_2$ we get: $$(v,u_1)=s-2t+1=0 \\ (v,u_2)=2s-t=0$$ The solution of this linear system is $s=1/3$ and $t=2/3$. By substituting this in $v$ we get the direction vector of the line $r$ on which lies our segment. More conveniently let $u=3v=(-1, 1, 1)$, which represents the same direction. What remains to be found is a point $P=(x_0, y_0, z_0)\in r$ so that we can write

$$r= P+\langle u \rangle=\begin{pmatrix} x_0-t\\ y_0+t\\ z_0+t \end{pmatrix} $$ By imposing the intersections of $r$ with $\ell_1$ and $\ell_2$ you can explicit out $x_0,y_0,z_0$.

Answer 4

Consider the diagram of Eckhard. Project the entire cube vertically and let the projection of the blue line be $XY$ with $Y$ lying on the edge of the face. $X$ lies on the red line of the top face. $XY$ is perpendicular to the red line of the top face. Thus $XY$ makes an angle of $\frac {\pi}{4}$ with the edge. Draw a line from $X$ perpendicular to the edge containing $Y$ and a horizontal line from $X$ to the edge perpendicular to the edge containing $Y$. Label these two points $P$ and $Q$ respectively. Note that $XP = XQ = PY$ and by symmetry (consider projections in the front face), if $D$ is the leftmost vertex on the edges containing $Y$, $YD = XP$ as well. It follows that $XP = \frac{1}{3}$. By looking at the projection in the front face, we see that the required length is $\sqrt{\frac{1}{3^3}+\frac{1}{3^2}+\frac{1}{3^2}} = \frac{1}{\sqrt{3}}$